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pishuonlain [190]
3 years ago
5

A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo

ut 61.0% of the roof. A typical annual rainfall is about 14.0 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water in a full tank of that size weigh (in N and in lbf)?
Engineering
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

volume  = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

volume  =  A'\times H

where A' = 61% of A

              = 0.61\times 2667 = 1626.87 sq ft

volume  =  1626.87 \times (\frac{14}{12} ft)

               =1898.015 ft^3

in\ m^3 = \frac{ 1898.015}{35.315} =   53.7457 m^3

in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon

weight = \rho Vg

       = 1000\times 53.74\times 9.8

             =526652 N

In\ lbf =  \frac{526652}{4.448} = 118396.08 lbf

You might be interested in
An air standard cycle with constant specific heats is executed in a closed system with 0.003 kg of air and consists of the follo
Vsevolod [243]

Answer:

a) Please see attached copy below

b) 0.39KJ

c)  20.9‰

Explanation:

The three process of an air-standard cycle are described.

Assumptions

1. The air-standard assumptions are applicable.

2. Kinetic and potential energy negligible.

3. Air in an ideal gas with a constant specific heats.

Properties:

The properties of air are gotten from the steam table.

b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.

P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K

T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K

Qin=m(u₂₋u₁)=mCv(T₂-T₁)

=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ

Qout=m(h₃₋h₁)=mCp(T₃₋T₁)

=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ

Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ

c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰

7 0
2 years ago
A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The
Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

       P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

      h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

       s1 = s-P1 = 0.6492 KJ/kg.K

       v1 = v-P1 = 0.001010 m^3 / kg

       

       P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

       s2 = s1 = 0.6492 KJ/kg.K   .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

   

                           w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}

- From the following relation we can determine ( h2 ) as follows:

                          h2 = h1 + wp

                          h2 = 191.81 + 8.0699

                          h2 = 199.88 KJ/kg

                           

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

              P3 = 8 MPa

              T3 = ?  ( assume fluid exist in the saturated vapor phase )

              h3 = hg-P3 = 2758.7 KJ/kg

              s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

                          q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

             P4 = 10 KPa

             s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

             

- Compute the quality of the mixture at condenser inlet by the following relation:

                           x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947

- Determine the isentropic ( h4s ) at this state as follows:

                          h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}        

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

                         h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

                        w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

                       W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

                        n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31

Answer: The thermal efficiency of the cycle is 0.31

       

   

7 0
3 years ago
During a medical evaluation, the doctor can __________.
Elan Coil [88]

Answer:

Treat the patient

i hope this is ur answer

8 0
2 years ago
Common car loan duration
chubhunter [2.5K]

Answer:

In 2019, the average term length was 69 months for new cars and 65 months for used vehicles. Most car loans are available in 12 month increments, lasting between two and eight years. The most common loan terms are 24, 36, 48, 60, 72, and 84 months, according to Autotrader

Explanation:

4 0
2 years ago
In particular, a system may or may not be (1) Memoryless, (2) Time invariant, (3)Linear, (4) Casual, (5) Stable.
egoroff_w [7]

Answer:

a. True

Explanation:

A system may be sometimes casual, time invariant, memoryless, stable and linear in particular.

Thus the answer is true.

A system is casual when the output of the system at any time depends on the input only at the present time and in the past.

A system is said to be memoryless when the output for each of the independent variable at some given time is fully dependent on the input only at that particular time.

A system is linear when it satisfies the additivity and the homogeneity properties.

A system is called time invariant when the time shift in the output signal will result in the identical time shift of the output signal.

Thus a system can be time invariant, memoryless, linear, casual and stable.

4 0
2 years ago
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